Question

The line 3y+x=25 is a normal to the curve y=x2-5x+k.find the value of constant k.

Answer 1

Answer:

k = 11.

Step-by-step explanation:

y = x^2 - 5x + k

dy/dx = 2x - 5 = the slope of the tangent to the curve

The slope of the normal = -1/(2x - 5)

The line  3y + x =25 is normal to the curve so finding its slope:

3y = 25 - x

y = -1/3 x + 25/3 <------- Slope is -1/3

So at the point of intersection with the curve, if the line is normal to the curve:

-1/3 = -1 / (2x - 5)

2x - 5 = 3  giving x = 4.

Substituting for x in y = x^2 - 5x + k:

When x = 4, y =  (4)^2 - 5*4 + k  

y = 16 - 20 + k

so y = k - 4.

From the equation y = -1/3 x + 25/3,  at x = 4

y = (-1/3)*4 + 25/3 = 21/3 = 7.

So y = k - 4 = 7

k = 7 + 4 = 11.